Solve ∫ 2*91cos(-3*07x-4*53132)+3*09 wrt x

$$\int 2 \cdot 91 \cos ( - 3 \cdot 07 x - 4 \cdot 53132 ) + 3 \cdot 09 d x$$

Evaluate

$$182\cos(212528)x+С$$

Solution Steps

Do the multiplications.

$$\int 182\cos(-3\times 0\times 7x-4\times 53132)+0\times 9\mathrm{d}x$$

Multiply -3 and 0 to get 0.

$$\int 182\cos(0\times 7x-4\times 53132)+0\times 9\mathrm{d}x$$

Multiply 0 and 7 to get 0.

$$\int 182\cos(0x-4\times 53132)+0\times 9\mathrm{d}x$$

Anything times zero gives zero.

$$\int 182\cos(0-4\times 53132)+0\times 9\mathrm{d}x$$

Multiply 4 and 53132 to get 212528.

$$\int 182\cos(0-212528)+0\times 9\mathrm{d}x$$

Subtract 212528 from 0 to get -212528.

$$\int 182\cos(-212528)+0\times 9\mathrm{d}x$$

Multiply 0 and 9 to get 0.

$$\int 182\cos(-212528)+0\mathrm{d}x$$

Anything plus zero gives itself.

$$\int 182\cos(-212528)\mathrm{d}x$$

Find the integral of 182\cos(-212528) using the table of common integrals rule \int a\mathrm{d}x=ax.

$$182\cos(-212528)x$$

Simplify.

$$182\cos(212528)x$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$182\cos(212528)x+С$$

Differentiate w.r.t. x

$$182\cos(212528)$$